Problem: Find the gradient of $f(x, y) = \ln(x) + y^2$ at $(1, 3)$. $\nabla f(1, 3) = ($
Explanation: The gradient of a scalar field is all its partial derivatives put together into a vector. For a 2D scalar field, this looks like $\nabla f = (f_x, f_y)$. Let's find $f_x$ and $f_y$. $\begin{aligned} f_x &= \dfrac{\partial}{\partial x} \left[ \ln(x) + y^2 \right] \\ \\ &= \dfrac{1}{x} \\ \\ f_y &= \dfrac{\partial}{\partial y} \left[ \ln(x) + y^2 \right] \\ \\ &= 2y \end{aligned}$ Now we can evaluate the partial derivatives we found at the point $(1, 3)$. $\begin{aligned} f_x(1, 3) &= \dfrac{1}{x} = \dfrac{1}{1} = 1 \\ \\ f_y(1, 3) &= 2y = 6 \end{aligned}$ The gradient of $f$ at $(1, 3)$ is $\nabla f = (1, 6)$.